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If the time period $t$ of the oscillation of a drop of liquid of density $d$, radius $r$, vibrating under surface tension $s$ is given by the formula $t=\sqrt{r^{2b}s^{c}d^{a/2}}.$ It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}.$ The value of $b$ should therefore be :
If the time period $t$ of the oscillation of a drop of liquid of density $d$, radius $r$, vibrating under surface tension $s$ is given by the formula $t=\sqrt{r^{2b}s^{c}d^{a/2}}.$ It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}.$ The value of $b$ should therefore be :
Updated On: Jul 4, 2024
- $\frac{3}{4}$
- $\sqrt{3}$
- $\frac{3}{2}$
- $\frac{2}{3}$
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The Correct Option is C
Solution and Explanation
Answer (c) $\frac{3}{2}$
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